Chapter 7:
Kinetic Energy & Work
Lecture 15
9/30/09
Kinetic Energy & Work
Goals for this Lecture:
Discuss Work done by the gravitational force
Discuss Work done by the spring force
Discuss Work done by a non-constant force
Introduce the concept of Power
Work By Gravity
The force of gravity always points down.
Case 1:
Case 2:
d
!=180°
!
Fg = mg
W = F·d = -mgd
Gravity slows
down the object
d
!=0°
!
Fg = mg
W = F·d = +mgd
Gravity speeds up
the object
Case 3:
d
h=dcos !
!
!
Fg = mg
W = F·d = mgdcos!
= mgh
Lifting/Lowering A Book
What is the work done in lifting (or dropping) a wide
receiver (m) from the ground to shoulder height (h)
(shoulder to ground)?
Hint: WM. Wright = FM. Wright·h will not work because the
force being exerted on the wide receiver is not
necessarily constant
Lifting/Lowering A Book
What is the work done in lifting (or dropping) a wide
receiver from the ground to shoulder height
(shoulder to ground)?
Let’s use the Work-Kinetic Energy Theorem.
Case 1: Lifting
"K = Wtot = Wgrav. + WM. Wright
"K = Kf - Ki = Wgrav. + WM. Wright
0 - 0 = Fgrav.·h + WM. Wright
(Fgrav = constant)
0
= -(mg)h + WM. Wright, WM. Wright = +mgh
Case 2: Dropping (i.e. setting down gently)
"K = Kf - Ki = Wgrav. + WM. Wright
0 - 0 = Fgrav.·h + WM. Wright
(Fgrav = constant)
0
= +(mg)h + WM. Wright, WM. Wright = -mgh
Work & Variable Forces
How do we calculate the work done by a
non-constant force?
Consider a force that varies between
xi and xf as shown in the graph
Work & Variable Forces
How do we calculate the work done by a
non-constant force?
Divide up the path into many (N) small regions
with “almost constant” force.
The work done in each interval j is !Wj = Fj·!x
Add up all the contributions: W = "Wj = " Fj·!x
Work & Variable Forces
How do we calculate the work done by a
non-constant force?
Let the length of each interval !x -> 0
W = lim "Wj = " Fj·!x = #F(x)·dx
i.e. Work is the area under the curve
Work-Kinetic Energy Theorem
Change in the kinetic energy of an object is
equal to the net work done on the object
"K = WTOT
Is this still true for a variable force?
W = #F(x)·dx = #ma(x)·dx = m#(dv/dt)·dx
= m#dv·(dx/dt) = m#dv·(v) = #m(vf2 - vi2)
= "K
Yes
Work By a Spring Force
A spring that is displaced from its equilibrium
length exerts a force that is:
Direction: opposite to the direction of displacement
from the equilibrium point
Magnitude: linearly proportional to the displacement
!
! (Hooke’s Law)
F = - kd
k is the spring constant:
Large k = hard spring
Small k = soft spring
As a theorist, Robert Hooke (1635-1703) is mostly
remembered for arguing with Isaac Newton over the nature
of light and gravity, a long-running debate that is said to
have left both men forever bitter toward each other.
Work By a Spring Force
Consider a spring that is stretched from an
initial position xi to a final position xf by an
external force (i.e. you hand), what is work done
by the spring?
Wspring =
Fspring(x)·dx = -kx·dx = - k(x f2 - xi2)
If xi = 0, Wspring = - kx f2
Recall: K = WTOT
K = 0 = WTOT + Whand, so
Whand = + k(x f2 - xi2)
i.e. Your hand has to do
positive work to stretch an object on a spring while
the spring does negative work.
Power
Power is the rate at which Work is being done by a
given Force:
P = dW/dt (instantaneous power)
Over a finite time interval we can use average power:
Pavg = !W/!t
Units: The unit of Power is the Watt (W)
P = !W/!t -> [Watt] = [J] / [s]
P·t = Work = Energy. The kilowatt-hour
[kWh] is a unit of work. It is used
because many engines are readily
characterized by their power output, and
how long they are running. kWh is the
common unit used by electrical utilities
(but it is really just Energy)
Power
Consider the earlier example of a force
acting on a particle:
The rate at which this
force does work on the
particle is:
v
P = dW/dt = (Fcos! dx)/dt
!
= Fcos! dx/dt = Fvcos! = !
F·v
What is the power exerted by the centripetal
force on object moving along a circular path?
Spring Power
A block of mass m is
attached to a spring with spring constant k. The
block is pulled to a distance +xmax from
equilibrium and released. A time T (seconds) later
the block reaches the point of maximum
compression xmin.
What is the average power exerted by the spring
on the block?
Pavg = !W/!t = !(KE)/!t = (0-0)/T = 0 J/T s = 0 W
Spring Power
A block of mass m is
attached to a spring with spring constant k. The
block is pulled to a distance +xmax from
equilibrium and released. A time T/2 (seconds)
later the block passes through the equilibrium
point xeq=0.
What is the average power exerted by the spring
on the block?
Pavg = W/ t
W = -kx·dx = - k(x f2 - xi2) = - k(0 - x max2)
= kx max2
Pavg = W/ t = kx i2 / (T/2) = kx max2/T Watts